Essentials of Gregor Mendel Biology - Green River … · Essentials of Biology ... 10.1 Mendel’s Laws Gregor Mendel ... Explain why the expected phenotypic ratio is different than

Essentials of

Biology

Sylvia S. Mader

Chapter 10

Patterns of Inheritance

Lecture Outline

•Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

10.1 Mendel’s Laws

Gregor Mendel• Austrian monk

• Worked with pea plants in 1860’s

• First person to analyze patterns of inheritance

• Deduced the fundamental principles of genetics.

Mendel working in his Abbey Garden

• Studied garden peas

• Easily manipulated

• Can self-fertilize

• Many traits controlled by only one gene

•Figure 10.1 •a. Flower structure

•ovary

•stigma

•anther

•Pollen grains containing sperm are•produced in the anther . When pollen•grains are brushed onto the stigma,•sperm fertilizes eggs in the ovary .•Fertilized eggs are located in ovules,•which develop into seeds.

•Figure 10.2 •Garden pea anatomy & traits

• Used garden pea, Pisumsativa

Easy to cultivate, short generation time

Normally self-pollinate but can be cross-pollinated by hand

• Mendel’s Experimental Procedure

•Figure 10.2 (cont.)

•Cut away anthers.

•b. Cross pollination

•1

•2

•3

•Brush on pollen from•another plant.

•The results of cross from a•parent that produces round,•yellow seeds × parent that•produces wrinkled•yellow seeds.

• Used true-breeding varieties

offspring were like the parent plants and each other

• Kept careful records of large number of experiments

• Used mathematical laws of probability to interpret results to develop his theory of inheritance


• Parents• (P)

• Offspring• (F1)


• Used true-breeding varieties

offspring were like the parent plants and each other

• Kept careful records of large number of experiments

• Used mathematical laws of probability to interpret results to develop his theory of inheritance

•Cross between true breeding long and short pea plants

•F1

Generation:

•All long

•Long •Short

•P Generation

•Which phenotype is dominant?•Recessive?

•F2 Phenotypic Ratio: ¾ of offspring are long

• (3 Long : 1 Short) ¼ of the offspring are short

•F2 Long Long Long Short

•Self-fertilization of F1

•How did Mendel Explain these results?

•Nucleus

•Long stem•allele

•Short stem•allele

•Purple flower•allele

•White flower•allele

•Flower color gene:

•A pair of•homologous

•chromosomes

•A pair of•homologous

•chromosomes

•Stem length gene:(a) Homozygous dominant

(two matchingdominant alleles)

(b) Heterozygous (nonmatchingalleles)

Phenotype:Long Long Short

Long stemallele

Long stemallele

A pair ofhomologouschromosomes

Genotype:

Long stemallele

Short stemallele

Short stemallele

Short stemallele

Genotype: (c) Homozygous recessive(two matching recessive alleles)

Heterozygous plants

Female Male

Long Long

(b) Alleles segregatein meiosis

Genotypeof eggs

Genotypeof sperm

A Monhybrid Cross:

Ll x Ll

Genotype of Eggs Genotype of Sperm

Alleles combine randomly during fertilization

PunnettsquareGenotypic ratio:

1 L L : 2 L l : 1 l l

Phenotypic ratio:

3 long : 1 short

Mendel’s Law of Segregation

• Alleles separate from each other during meiosis

• Results in gametes with one or the other allele, but never both

Formation of gametes from a pre-gamete cell

Genotype of Pre-gamete cell:

meiosis

Genotype of Gametes:

A a

•Aa

•½ •½

Genetic Terms

1. Phenotype Vs. Genotype—what’s the difference.

2. What are Alleles?

– Alternate forms of a gene… e.g.’s?

3. Where are the alleles of a gene located?

4. How many alleles can a person inherit for any one trait?

• How many alleles are there in a population for a particular trait such as

hair color?

5. What’s the relationship between alleles and homologous chromosomes?

6. Dominant vs. recessive alleles—what’s the difference? E.g.’s?

7. How can you determine if an allele is dominant or recessive?

• Genotype: alleles individual receives at fertilization

Homozygous: 2 identical alleles

• Homozygous dominant

• Homozygous recessive

Heterozygous: 2 different alleles

• Phenotype: physical appearance of individual

Determined by genotype and environmental factors

Genotype vs. Phenotype

•Figure 10.5 Alleles on Homologous chromosomes Homologous Pair of Chromosomes with 3 linked genes

Family Pedigrees

• Shows the history

of a trait in a

family from one

generation to

another

• Allows

researchers to

determine if a

phenotype is

dominant or

recessive

Dominant vs. Recessive Phenotype

Dominant Phenotype Recessive Phenotype

Genotype PhenotypeRR = _____________________ ___________________

Rr = _____________________ ___________________

rr = _____________________ ___________________

Allele Symbols:

R = Tongue Roller

r = Nonroller

Types of Genotypes and their resulting Phenotypes

Attached vs. Free Earlobes

1. Due to a recessive or

dominant allele?

2. Sex-linked or

Autosomal?

3. Must examine a

pedigree to answer

these questions

Marr Family Pedigree for Earlobe Attachment1. Which allele is

dominant?

Recessive?

• Is the allele for

earlobe

attachment sex-

linked (X-

linked) or

autosomal?

• What are the

genotypes of all

family

members?

•Marr Family Pedigree for Earlobe Attachment A family pedigree for Deafness

• Is deafness a dominant or recessive trait?

• How can you tell?

Allele Symbols: D = _____________________ d=_______________________

JoshuaLambert

AbigailLinnell

JohnEddy

HepzibahDaggett

AbigailLambert

JonathanLambert

ElizabethEddy

Female Male

Deaf

Hearing

A family pedigree for Deafness

• Is deafness a dominant or recessive trait?

• How can you tell?

Allele Symbols: D = _____________________ d=_______________________

•Dd•Joshua•Lambert

•Dd•Abigail•Linnell

•D_•John•Eddy

•D_•Hepzibah•Daggett

•D_•Abigail•Lambert

•dd•Jonathan•Lambert

•Dd•Elizabeth

•Eddy

•Dd •Dd •dd •Dd •Dd •Dd •dd

•Female •Male

•Deaf

•Hearing

Common Monogenic Human Traits

Dominant Allele

1. Free Earlobes

2. Straight Thumb

3. Long eyelashes

4. Normal health

5. Normal health

6. Normal R.B.C’s

7. Huntington’s Disease

Recessive Allele

Attached Earlobes

Hitchhiker’s thumb

Short eyelashes

Cystic Fibrosis

Tay-Sac’s Disease

Sickle cell anemia

Normal Health

Common Polygenic Human Traits

Dominant

1. Dark-colored hair

2. Curly hair

3. Dark eyes

4. Hazel or green eyes

5. Tall

6. Dark skin

Recessive

Light-colored hair

Straight hair

Light eyes (blue or gray)

Blue or gray eyes

Short

Light skin

Common Sex-linked Recessive Human Traits: X-linked

X-linked recessive traits

– Uncommon in females—why?

– Father must have disease and mother must be a carrier for a daughter to have the disease.

1. Color Vision

XN = Normal color vision; Xn = Red/Green Colorblind

2. Hemophilia:

XN = Normal blood clotting; Xn = bleeder

3. Duchenne Muscular Dystrophy

XN = Normal muscles; Xn = muscular dystrophy

Laws of Probability—application to inheritance

1. The results of one trial of a chance event do not affect

the results of later trials of that same chance event

• E.g. Tossing of a coin, gender of children, etc.

Laws of Probability—application to inheritance

2. The Multiplication Rule: The chance that two or more

independent chance events will occur together is equal

to the product of their chances occurring separately

a. What are the chances of a couple having 9 girls?

b. E.g. What are the chances of a couple having a boy with the

following characteristics:

– Brown hair (3/4), Non-tongue roller (1/4), Blue eyes (1/4), Attached

earlobes (1/4)

How to Solve Genetics Problems

Sample Problem: Mom and dad are heterozygous for tongue rolling where tongue rolling is dominant to non-rolling. What is the chance that the couple will produce a girl that is a non-roller?

Use the following steps as a general guide to solve this and other problems:

1. Select a letter to represent the gene involved

• Use upper case for the dominant allele, lower case for the recessive allele.

2. Write the genotypes of the parents.

3. Determine all possible gametes for each parent.

• Alleles for a trait segregate into separate gametes during meiosis

4. Determine the genotypes of the offspring.

• Make a Punnett square to represent all possible gamete combinations

between the two parents

5. Use the genotypes found in the Punnett Square to determine the possible

phenotypes of the offspring to answer the question.

Types of Genetics Problems

Monohybrid Crosses

• Involve only one trait such as …??

Sample Problem #1: True breeding parental pea plants were crossed to produce

the F1 generation, below. The F1 generation was inbreed to produce an F2

generation.

a.) Which allele is dominant? How do you know?

b.) Determine the genotypes and phenotypes for all 3 generations

c.) Predict the genotypic and phenotypic ratios for the F2.

P: Purple flowered pea plant x White flowered Pea Plant

F1: 100% Purple Flowered

F2: ???

A true breeding black mouse was crossed with a true breeding brown mouse to

produce the F1 generation, below. The F1 generation was then inbred to

produce an F2 generation.




P: Black mouse x Brown mouse

F1: 100% black mice

F2: ???

Monohybrid Cross Sample Problem #2

A mouse with black fur was crossed with a mouse with brown fur to produce

the F1 generation, below. The F1 generation was then inbred to produce the F2

generation. Dominance is the same as in sample problem #2.

a.) Determine the genotypes and phenotypes for all 3 generations

b.) Predict the genotypic and phenotypic ratios for the F2.

P: Black mouse x Brown mouse

F1: ½ black mice; ½ Brown

F2: ???


Use the information below to answer the following questions. Dominance is the

same as in the preceding problems involving mice.

a.) Calculate the phenotypic ratio of the F2.


c.) Determine the expected phenotypic and genotypic ratios for the F2.

d.) Explain why the expected phenotypic ratio is different than the actual

phenotypic ratio for the F2.

P: ??????????? x ??????????

F1: 100% black mice

F2: 27 Black mice + 10 Brown mice

Monohybrid Cross Sample Problem #4Answers to Sample Problem #1

True breeding parental pea plants were crossed to produce the F1 generation,

below. The F1 generation was inbreed to produce an F2 generation.


The purple allele is dominant to the white allele since the white phenotype does not appear in the F1.



P: Purple flowered pea plant = PP x White flowered Pea Plant =pp

F1: 100% Purple Flowered = Pp

F2: Genotypic Ratio: 1 PP : 2 Pp : 1 pp

Phenotypic Ratio: 3 Purple : 1 white

A true breeding black mouse was crossed with a true breeding brown mouse to

produce the F1 generation, below. The F1 generation was then inbred to



The black allele is dominant to the brown allele since the brown phenotype does not appear in the F1.



P: Black mouse = BB x Brown mouse = bb

F1: 100% black mice = Bb

F2: Genotypic Ratio: 1 BB : 2 Bb : 1 bb

Phenotypic Ratio: 3 Black : 1 Brown

Answers to Sample Problem #2

A mouse with black fur was crossed with a mouse with brown fur to produce

the F1 generation, below. The F1 generation was then inbred to produce the F2

generation. Dominance is the same as in sample problem #2.

a.) Determine the genotypes and phenotypes for all 3 generations

b.) Predict the genotypic and phenotypic ratios for the F2.

P: Black mouse = Bb x Brown mouse = bb

F1: ½ black mice = Bb; ½ Brown = bb

F2: ½ black mice = Bb; ½ Brown = bb


Use the information below to answer the following questions. Dominance is the

same as in the preceding problems involving mice.

a.) Calculate the phenotypic ratio of the F2.


c.) Determine the expected phenotypic and genotypic ratios for the F2.

d.) Explain why the expected phenotypic ratio is different than the actual

phenotypic ratio for the F2.

P: ??????????? x ??????????

F1: 100% black mice

F2: 27 Black mice + 10 Brown mice


A couple, Jack and Jill, is concerned about having a

child with cystic fibrosis. Although both of Jack’s and

both of Jill’s parents are healthy and show no signs of

cystic fibrosis, both Jack and Jill each had a sister die of

the disease. The couple went to a clinic to be genetically

tested for cystic fibrosis and were each found to be

heterozygous for cystic fibrosis. What are the chances of

Jack and Jill having a….

a.) phenotypically healthy child?

b.) child that is homozygous dominant?

Heterozygous? Homozygous recessive?

c.) girl with cystic fibrosis? Boy with cystic fibrosis?


Salty sweat due to

altered salt secretion

in sweat ducts

Lungs

Pancreas

Testis

Infertilty in

males due

to clogged

sex ducts

Problems with

digestion due

to clogged

duct from

pancreas

Mucus-clogged

airways

Symptoms of cystic fibrosis

Cell lining ducts

of the body

Gaucher disease is an autosomal recessive disorder.

What are the chances of a phenotypically normal and

healthy couple having a child with Gaucher disease if

each partner has a brother with GD and the parents of

the couple are phenotypically healthy?

Hints: (This problem is more complex than you may think!)

1.) Neither couple knows their genotype.

2.) Being phenotypically healthy eliminates one of

the possible genotypes for the couple.


Test Cross• Used to determine if an organism with the dominant phenotype

is homozygous dominant or heterozygous

• Involves the cross of an organism with the dominant phenotype

with __________________________________.

e.g. Free earlobes is dominant to attached earlobes in humans. How

could your instructor determine if he is homozygous or

heterozygous for free earlobes?

Instructor (Free Earlobes) X Wife (_____________________)

Daugher (_____________________)

Conclusion??

Types of Genetics Problems

Dihybrid Crosses

• Involve two traits. Such as …??

Dihybrid Cross Sample Problem #1: True breeding parental pea plants were

crossed to produce the F1 generation, below. The F1 generation was inbreed to


a.) Which alleles are dominant? How do you know?


P: Long & Purple flowered pea plant x Short & White flowered Pea Plant

F1: All Long & Purple Flowered pea plants

F2: 9 Long & Purple : 3 Long & White : 3 Short & Purple : 1 Short & White

In general, the F2 of a Dihybrid Cross: 9 D1 & D2 : 3 D1 & R2 : 3 R1 & D2 : 1 R1 & R2

Parents (P)

Long & purple

(double dominant) Short & white

(double recessive)

All long & purpleF1

Illustration of

the Dihybid

Cross in

Sample

Problem #1

(slide 1 of 2)

Genotypes of Sperm

F2 Phenotypic ratio:

F1 genotypes: LlPp

Genotypes of Eggs

Illustration of the Dihybid Cross in Sample Problem #1

(slide 2 of 2)

Explaining Dihybrid Crosses

Mendel’s Law of Independent Assortment

Each pair of alleles separates (segregates) independent

from other pairs of alleles during gamete formation

unless the genes for these alleles are found on the same

chromosome

Results in a 9 : 3 : 3 : 1 Phenotypic Ratio in the F2!

• Gene for earlobes and

hairline on different

chromosomes

• Gametes have all

possible combination

of alleles.

•Figure 10.8 Mendel’s laws and meiosis

Figure 10.8 Potential gametes produced by a person who is EeWw

•E •E

•W •W

•E •E

•E

•W

•W

•E

•W

•W

•e •e

•w •w

•e •e

•e

•w

•w

•e

•w

•w

•E •E

•w •w

•w •w

•E •E

•E •E

•e •e

•e •e

•W •W

•W •W

•w •w

•Meiosis I

•Meiosis II

•either •or

•EW •ew •Ew •eW

•E

•W

•w

•e

•W •W

•e •e

•one•pair

•one•pair

•Key:

•w = straight hairline

•E = unattached earlobes

•e = attached earlobes

•W = widow’s peak

•Parent cell with 2 pairs•of homologues.

•Homologues can•align either way•during metaphase I.

•All possible combinations•of chromosomes and•alleles result.

Determining Gametes for traits that assort independently

Traits that assort independently are on different homologous

pairs of chromosomes—I.e. the traits are not linked.

Number of genetically different gametes possible = 2n (where

n = the number of heterozygous traits)

Practice Problems

» How many genetically unique gametes are possible for the following

genotypes? List the genotypes of all possible gametes for #’s 1-5, below.

1. AaBb

2. AABb

3. AABBCC

4. AaBbCc

5. AaBBCc

6. AaBbCcddEe

Using the Probability Method to Solve “Multi-hybrid” Problems

From the crosses below, what are the chances of

producing an organism with all

» dominant phenotypes?

» recessive phenotypes?

» homozygous dominant genotypes?

1. AaBb x AaBb

2. AaBbCc x AaBbCc

3. AaBBCc x aabbcc

How to use the probability method

1. Treat the problem as if it consisted of several monohybrid crosses

2. Determine the gametes for each of these monohybrid crosses

3. Make a Punnett square for each of the monohybrid crosses

4. Use the information from each Punnett square and the

“multiplication rule” to solve the problem

Analysis of Pedigrees1. Is the disease dominant or recessive? How can you tell?

2. Autosomal or Sex-linked inheritance? How can you tell?

3. Can you determine the genotypes of all individuals?

» For which phenotype do we always know the genotype?

Pedigree #1 (Purple shading indicates genetic disease)

Analysis of Pedigrees

1. Is the disease dominant or recessive? How can you tell?



» For which phenotype do we always know the genotype?

Pedigree #2 (shading indicates genetic disease)

Human Polydactyly: Extra Finger or Toe

1. Is the disease dominant or recessive? How can you tell?



Human Polydactyly

I

II

III

IV

Analysis of British Royal Family Pedigree1. Is the disease dominant or recessive? Autosomal or sex-linked?

2. Determine genotypes

10.2 – 10.3 Beyond Mendel’s Laws

Sex-linked recessive inheritance

» Recessive on X-chromosome

– e.g. Hemophilia, colorblindness, Androgen Insensitivity Syndrome (e.g.

Jamie Lee Curtis?) http://www.medhelp.org/www/ais/ ;

http://www.youtube.com/watch?v=ETIxoQGVjos

Incomplete dominance

» e.g. Snapdragons

– red flower x white flower pink flower

» Sickle cell anemia

NN = healthy; nn = sickle cell anemia (deadly); Nn = sickle cell trait

Co- dominance: ABO Blood Groups

» Blood types: A, B, AB, O

Is it possible to be XY and female? XX and male? The Maria Patino Story

Maria Patino couldn't sleep before her 1st race at the 1985 World University Games in Japan. She was the Spanish National Champion and scheduled to perform in the 60m hurdles the next day but she wasn't sure if she would be able to compete. Earlier that day she reported to "Sex Control" which scraped cells from her cheek to test for sex chromosomes. She had passed the test in 1983 in Helsinki but had forgotten to bring her "Certificate of Femininity". A few hours after the test officials told her the test was abnormal but not to worry. But she worried all night. Did she have leukemia that killed her brother? Did she have AIDS? The next morning they did a follow up check and she failed the sex test! She had male sex chromosomes, XY! Sports officials decided Maria should fake an injury in warm-up so no one would suspect why she withdrew. Spanish officials told her she had to drop out of sports.

Maria was aghast:» “I knew I was a women in the eyes of medicine, God and most of all, in my own eyes.”

» It came out in the newspapers. Her boyfriend left her and other friends also. Spanish sports officials took her records out of the record books.

Marias phenotype: female genitalia, female body proportions, sexually attracted to males, but no uterus, sterile and no pubic hair.

Let's investigate how sex is determined to try to figure out what is happening.

http://www.medhelp.org/www/ais/

http://www.youtube.com/watch?v=ETIxoQGVjos

Human Sex Chromosomes

Sex chromosomes in humans

» Female Genotype = XX

» Male Genotype = XY

Sex-linked Alleles are carried on the X-chromosome

» ~1000 genes on X-chromosome

Y-chromosome

» Only ~20 genes on Y-chromosome

– Mostly involved with male fertility

» SRY gene on Y chromosome activated around the 7th week of pregnancy

– Gene product stimulates gonads to differentiate into male sex organs.

– SRY = Sex-determining Region, Y-chromosome

Effect of SRY Gene Activation on the

Development of the Internal Sex

Organs

Effect of SRY Gene Activation

on the Development of the External Sex

Organs

Pedigree of a family with varying degrees of Androgen Insensitivity

Holterhus, P. et al. 2000. Journal Clinical Endocrinol Metab. 85: 3245-3250 (Avail. http://jcem.endojournals.org/cgi/content/full/85/9/3245)

A family with four affected individuals, three brothers (B1–3) and their uncle, displaying strikingly different external genitalia.

•B1: ambiguous •B2: severe micropenis •B3: slight micropenis

•Uncle: micropenis & penoscrotal hypospadias.

•Abstract

•Molecular causes of phenotypic diversity in androgen insensitivity syndrome,occurring even in the same family, have rarely been identified. We report on a family with four affected individuals, three brothers (B1–3) and their uncle, displaying strikingly different external genitalia: B1, ambiguous; B2, severe micropenis; B3, slight micropenis; and uncle, micropenis and penoscrotal hypospadias. All had been assigned a male gender. We detected the same L712F mutation of the androgen receptor (AR) gene in each subject. Methyltrienolonebinding on cultured genital skin fibroblasts of B2 suggested moderate impairment of the ligand-binding domain [maximal binding capacity, 38.2 fmol/mg protein (normal); Kd, 0.21 nmol/L; normal range, 0.03–0.13 nmol/L]. In trans-activation assays, the mutant 712F-AR showed considerable deficiency at low concentrations of testosterone (0.01–0.1 nmol/L) or dihydrotestosterone (0.01 nmol/L). Remarkably, this could be fully neutralized by testosterone concentrations greater than 1.0 nmol/L. Hence, the 712F-AR could switch its function from subnormal to normal within the physiological concentration range of testosterone. This was reflected by an excellent response to testosteronetherapy in B1, B2, and the uncle. Taking into account the well documented individual and time-dependent variation in testosterone concentration in early fetal development, our observations clearly illustrate the potential impact of varying ligand concentrations for distinct cases of phenotypic variability in androgen insensitivity syndrome.

Normal Karyotype of Human Chromosomes

What are homologous chromosomes?

What gender?

Sex vs. autosomal chromosomes?

Sperm

ParentsMaleFemale

Chromosomessegregate in

meiosis

Offspring

Eggs

Twodaughters

Twosons

Gender

Determination

in Humans

• Gender is determined by the presence of the SRY-gene on the Y-chromosome

• SRY gene is turned on around the 7th week of gestation.

• XY females = Androgen Insensitivity Syndrome

– X-linked recessive

– Androgen receptor doesn’t recognize testosterone

– Consequences?

Test for Red-Green Colorblindness• Colorblindness is caused

by a malfunction of light-sensitive cells in the retina of the eyes

• What number do your see?

• Like all X-linked recessive traits, colorblindness is very rare in women. Why??

Sperm

Parents

Normal MaleCarrier Female

Offspring

Eggs

XN Xn XN Y

XN Y

Xn

XN

XN Xn

XN XN

XnY

XN Y

Colorblindness:

X-linked Recessive

Inheritance

2 Healthy daughters

1 healthy son +

1 color blind son

Carrier

Female

XN Xn

Normal

Male

XN Y

X

+

Pedigree Duchenne Muscular Dystrophy

• Mode of inheritance?

• Autosomal or Sex-linked?

• Genotypes?

Sample Problem

Mary’s paternal and maternal grandfathers

are both colorblind. There is no evidence of

colorblindness in either grandmother’s

family histories.

a. What is Mary’s genotype? Phenotype?

b. What are the chances that Mary’s brother is

colorblind?

X-Linked Dominant ExampleCongenital Bilateral Ptosis: Droopy Eyelids

Locus: Xq24-Xq27.1

C.Stern et al. (1964) Am J Hum Gen. 16:467.

Hairy ears, Y-LINKED?

HYPERTRICHOSIS PINNAE AURIS

Incomplete Dominance

The dominant allele is incompletely dominant

over the recessive allele

Phenotype of heterozygous individuals is in-

between that of the homozygous dominant and

homozygous recessive phenotypes

» E.g. Snapdragons, sickle cell anemia

Incomplete Dominance in Snapdragons

Sickle Cell Anemia—an example of incomplete dominance

Normal Red Blood Cells Sickled R.B.C.’s clump together

and clog blood vessels

• Uncommon in U.S.A. (~1 in 60,000)

• Common in West Africa (~1 in 50) and African Americans (~1 in 400)

– Lethal in the homozygous recessive condition

– What is the adaptive value of heterozygous condition in West Africa?

P: Male with Sickle Cell Trait (Hh) x Female with Sickle Cell Trait (Hh)

Gametes of Female with

sickle cell trait

Gametes of Male with sickle cell trait

H = healthy hemoglobin allele

h = sickle cell allele

H

H h

h

HH =

normal RBC

Hh =

RBC sickle when

levels are O2 Low

hh =

Sickled RBC’s

Hh =

RBC sickle when

levels are O2 Low

Sickle Anemia Pedigree: An example of incomplete dominance

Unaffected

Sickle cell trait

Sickle cell anemia

Decreased

Individual homozygousfor sickle-cell allele

Sickle-cell (abnormal) hemoglobin

Abnormal hemoglobin crystallizes,causing red blood cells to become sickle-shaped

Sickled cells

Breakdown ofred blood cells

Accumulation ofsickled cells in spleen

Physicalweakness

AnemiaHeartfailure

Pain andfever

Braindamage

Damage toother

organs

Clumping of cellsand clogging of

small blood vessels

Spleendamage

Impairedmental

function

ParalysisPneumoniaand otherinfections

Rheumatism Kidneyfailure

Pleiotrophy

• The impact of a

single gene on

more than one

characteristic

• Examples of

Pleiotrophy:

– Sickle-Cell

Anemia

– Gaucher Disease

– Cystic Fibrosis

Codominance: Blood Types

• Alleles

IA = Allele for Type A

IB = Allele for Type B

i = Allele for Type O

• IA is dominant to i

• IB is dominant to i

• IA and IB are codominant

• What do these alleles code for?

• How many alleles can you inherit?

Blood

Type

(Phenotype)

Surface

Molecule

on R.B.C.

Possible

Genotypes

A IAIA or IAi

B IBIB or IBi

AB IAIB

O ii

ABO Type Rh Type How Many Have It

O positive 37.4%44%

O negative 6.6%

A positive 35.7%42%

A negative 6.3%

B positive 8.5%10%

B negative 1.5%

AB positive 3.4%4%

AB negative .6%

•ABO and Rh Blood TypeFrequencies in the United States

(Source: Stanford School of Medicine: Blood Center)

Blood Types: Sample Problem #1

A couple has the type A and Type B, respectively.

Is it possible for them to have a child with the

following blood types? If so, what is the genotype

of each parent?

a. Type O

b. Type A

c. Type B

d. Type AB


A couple has the type A and Type AB,

respectively. Is it possible for them to have a child

with the following blood types? If so, what is the

genotype of each parent?

a. Type O

b. Type A

c. Type B

d. Type AB

http://bloodcenter.stanford.edu/about_blood/blood_types.html

Rhesus Factor—a RBC surface molecule

Phenotype Possible

Genotypes

Rh +(Rh positive)

RR or Rr

Rh-(Rh negative)

rr

Rh factor is inherited independently from the ABO system

Rh positive people:

» R.B.C’s have the Rhesus factor surface molecule

Rh Negative people:

» R.B.C’s w/o the Rhesus factor surface molecule

Alleles

» R = Rh factor

» r = no Rh factor


A couple has the type A+ and Type AB+,

respectively. What are the chances of the couple

having a child with the following phenotypes.

a. Type O+ b. Type O-

c. Type A+ d. Type A-

e. Type B+ f. Type B-

g. Type AB+ h. Type AB-

• Polygenic inheritance

Trait is governed by 2 or more sets of alleles.

Each dominant allele has a quantitative effect

on phenotype and effects are additive.

Result in continuous variation – bell-shaped

curve

Multifactorial traits – polygenic traits subject to

environmental effects

• Cleft lip, diabetes, schizophrenia, allergies, cancer

• Due to combined action of many genes plus

environmental influences

•Figure 10.11 Height in humans, a polygenic trait

•62 •64 •66 •68 •70 •72 •74

•Height in Inches•short •tall

•few •few

•Nu

mb

er

of

Me

n

•most•are•this

•height


•Courtesy University of Connecticut, Peter Morenus, photographer

• Environment and the phenotype

Relative importance of each can vary.

Temperature can effect coat color.

• Rabbits homozygous for ch have black fur where

the skin temperature is low.

• Enzyme encoded by gene is active only at low

temperatures.

•Figure 10.12 Coat color in Himalayan rabbits


10.4 Inheritance of Linked Genes

• Some fruit fly crosses violated the law of

independent assortment.

Offspring simply resembled one of the

parents.

• 2 traits on same chromosome – gene

linkage

• 2 traits on same chromosome do NOT

segregate independently.

• Recombination between linked genes

Linked alleles stay together –

heterozygote forms only 2 types of

gametes, produces offspring only with 2

phenotypes.

•Figure 10.16 Linked alleles and crossing-over

•tetrad

•a. Linked alleles usually stay together

•G •G

•R •R

•g •g

•r •r

•G •G

•R •R

•g •g

•r •r

•G

•R

•G

•R

•g

•r

•g

•r

•sister•chromatids

•alleles•are linked

•resulting•daughter

•chromosomes

•Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. • Occasionally crossing-over produces

new combinations.

Nonsister chromatids exchange genes.

Recombinant gametes have a new

combination of alleles.

•Figure 10.16 continued

•tetrad

•b. Crossing-over results in recombination of alleles

•G •G

•R •R

•g •g

•r •r

•G

•R

•g

•r

•G

•R

•g

•R

•G

•r

•g

•r

•nonsister•chromatids

•linked alleles•sometimes cross-over

•resulting•daughter

•chromosomes


• Distance between genes The closer 2 genes are on a chromosome, the less

likely they are to cross-over.

You can use the percentage of recombinant phenotypes to determine the distance between genes.

1% crossing-over = 1 map unit.

In a black-body and purple-eye cross, 6% of offspring are recombinant = genes are 6 map units apart.

Results can make a chromosome map.

•Figure 10.17 Linked alleles do not assort

independently

•P generation

•Offspring •Predicted •Observed

•25% •47%

•25% •47%

•25% •3%

•25% •3%

•1 gray body, red eyes

•1 black body, purple eyes•1 gray body, purple eyes

•1 black body, red eyes

• F1 Phenotypic Ratio

•G = gray body

•g = black body

•R = red eyes

•r = purple eyes

•Key:

•GgRr

•ggrr

•Ggrr

•ggRr

•ggrr•GgRr


•Figure 10.18 Mapping chromosomes

•6 map units •12.5 map units

•18.5 map units

•black•body

•purple•eyes

•vestigial•wings


Why are Calico Cats females, not males?

Genes for fur color in cats on the X chromosome:

XB = Black

Xb = Yellowish-orange

Calico cats are heterozygous: XBXb

» Why calico and not black?

» Due to X-inactivation—what’s that?

Gene for white is on an autosomal chromosome and unrelated to the alleles on the X-chromosome

X-inactivation in females

Anhidrotic Ectodermal DysplasiaCalico Cats: XBXb

Classic example of X-inactivation

Different fur producing cells randomly inactivate one of the X

chromosomes

» Happens during embryonic development

Gives the patchy calico fur pattern:

» Black patches have cells with the XB chromosome active

» Yellow patches have cells with the Xb chromosome active

Genotype XBXB XBXb XbXb XBY XbY

Phenotype

Documents

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